[過去ﾛｸﾞ] 現代数学の系譜 カントル 超限集合論2 (1002ﾚｽ)
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553(4): 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE 2020/03/20(金)11:34 ID:+qJdNaLm(5/8) AAS
＜転載＞
0.99999……は１ではない その７
2chｽﾚ:math
(抜粋)
79 2020/03/20(金) ID:WMaa4Quj
conglomerabilityの定義を理解した上でPrussの論文を読み直せば、
自説がPrussによって真正面から否定されてると理解できます
80 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE 2020/03/20(金) ID:+qJdNaLm
おサルさん、DR Pruss氏は、mathoverflowの彼の回答の前段で、conglomerabilityを出しているが
（下記引用ご参照）
最後は、”the function is measurable”が不成立だから、”dumb（ダメダメ） strategy”と言っているよ
（下記の通り）
QED
（＾＾；
（参考）
外部ﾘﾝｸ:mathoverflow.net
Probabilities in a riddle involving axiom of choice Denis氏 Dec 9 '13
DR Pruss氏
(抜粋)
By a conglomerability assumption, we could then conclude that P(X<=Y)=0, which would be absurd as the same reasoning would also show that P(Y<=X)=0.
In general, Mj will be nonmeasurable (one can prove this in at least some cases). We likewise have no reason to think that M is measurable. But without measurability, we can't make sense of talk of the probability that the guess will be correct.
That's a fine argument assuming the function is measurable. But what if it's not?
So there is an extension P′ of P such that P′-almost surely the dumb strategy works. Just let P′ be an extension on which the set of representatives has measure 1 and note that the dumb strategy works on the set of representatives.
外部ﾘﾝｸ:www.mdpi.com
Symmetry and the Brown-Freiling Refutation of the Continuum Hypothesis
by Paul Bartha
Symmetry 2011, 3(3), 636-652;
554(2): 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE 2020/03/20(金)12:00 ID:+qJdNaLm(6/8) AAS
>>553
DR Pruss氏は下記で、conglomerabilityの正確な意味がいまいち分からんけど
要するに”nonmeasurable”で、測度論的確率から外れているということでしょう （＾＾；
外部ﾘﾝｸ:en.wikipedia.org
Alexander Robert Pruss (born January 5, 1973) is a Canadian mathematician, philosopher, Professor of Philosophy and the Co-Director of Graduate Studies in Philosophy at Baylor University in Waco, Texas.
Pruss graduated from the University of Western Ontario in 1991 with a Bachelor of Science degree in Mathematics and Physics.
After earning a Ph.D. in Mathematics at the University of British Columbia in 1996 and publishing several papers in Proceedings of the American Mathematical Society and other mathematical journals,[4] he began graduate work in philosophy at the University of Pittsburgh.
外部ﾘﾝｸ:books.google.co.jp
Infinity, Causation, and Paradox
著者: Alexander R. Pruss
（P76-77 に conglomerabilityの説明があるが、正確な定義は分からないが、
P76に”But typically, where there is no coutable additibity, there is lack of conglomerability(Scervish,Seidenfeld,and Kanade 1984).”
と記されているので、”coutable additibity ”即ち σ-加法性 と密接に関連した（多分”σ-加法性”を拡張した）概念だと思う）
（更に附言すれば、現代の測度論的確率が、σ-加法性をベースに成立っているとすれば、DR Pruss氏の指摘は、要するに”nonmeasurable”で、測度論的確率から外れているということでしょう （＾＾； ）
558(3): 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE 2020/03/21(土)07:53 ID:gPebnXHG(1/13) AAS
>>557
おサルに分かるようには書けないなｗ
理解力の無いおサルには、正確に書いてもしかたないだろ？ｗ（＾＾；
mathoverflow（>>553）における 質問者 Denis氏に対する DR Pruss氏の回答が如し
つまり、DR Pruss氏は正確に回答しているが、質問者 Denis氏は ”the function is measurable”が理解できないみたい
” measurable”が分かってないんだな、質問者 Denis氏は
彼が、” measurable”に対する理解を示す発言皆無なんだよｗ
おサルは、それと同じだよｗ(゜ロ゜；
573(2): 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE 2020/03/21(土)20:28 ID:gPebnXHG(8/13) AAS
>>568
おサルさー、おまえ mathoverflowの DR Pruss氏議論が分かっていない 質問者 Denis氏そっくりの理解じゃんかｗ(゜ロ゜；
DR Pruss氏は、”That's a fine argument assuming the function is measurable. But what if it's not?”ってあるよね
で、質問者 Denis氏は、この議論には、全く入れなかった
ただ、壊れたレコードのように
”Our choice of index i is made randomly, but for this we only need the uniform distribution on {0,…,n}. It is made independently of the opponent's choice. ? Denis Dec 17 '13 at 15:21”
を繰返したのだった（＾＾；
（>>553より参考）
外部ﾘﾝｸ:mathoverflow.net
Probabilities in a riddle involving axiom of choice Denis氏 Dec 9 '13
DR Pruss氏
(抜粋)
By a conglomerability assumption, we could then conclude that P(X<=Y)=0, which would be absurd as the same reasoning would also show that P(Y<=X)=0.
In general, Mj will be nonmeasurable (one can prove this in at least some cases). We likewise have no reason to think that M is measurable. But without measurability, we can't make sense of talk of the probability that the guess will be correct.
That's a fine argument assuming the function is measurable. But what if it's not?
So there is an extension P′ of P such that P′-almost surely the dumb strategy works. Just let P′ be an extension on which the set of representatives has measure 1 and note that the dumb strategy works on the set of representatives.
外部ﾘﾝｸ:www.mdpi.com
Symmetry and the Brown-Freiling Refutation of the Continuum Hypothesis
by Paul Bartha
Symmetry 2011, 3(3), 636-652;
586(3): 現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE 2020/03/22(日)09:53 ID:TMbOZsnt(3/22) AAS
>>582
おサル、それ誤読だよ
”misunderstanding”は、下記引用の3)のとこでしょ
でも、面白いね、文献の”philosophical reason”の「 independently」の
”orthodox (Kolmogorovian) probability theory”と異なる見方（哲学だけれど）
（>>553より参考）
外部ﾘﾝｸ:mathoverflow.net
Probabilities in a riddle involving axiom of choice Denis氏 Dec 9 '13
DR Pruss氏
(抜粋)
show 6 more comments
1)Our choice of index i is made randomly, but for this we only need the uniform distribution on {0,…,n}. It is made independently of the opponent's choice. ? Denis Dec 17 '13 at 15:21
2)I was assuming that "independently" has the meaning it does in probability theory (P(AB)=P(A)P(B) and generalizations for σ-fields). But that does require a probabilistic description of the opponent's choice.
Of course, one could mean "independently" here in some non-mathematical causal sense. (And there may be philosophical reason for doing this: fitelson.org/doi.pdf )
Still, mixing the probabilistic with nonprobabilistic concepts might lead to some difficulties, though. ? Alexander Pruss Dec 18 '13 at 15:21
3)ah ok I see where the misunderstanding comes from, it's true that "independently" is ambiguous, because only one random variable is involved here.
But I think it still has a mathematical meaning in the sense "it does not depend on the opponent's choice", namely we have ∃x∀y where x is our strategy and y is our opponent's strategy (i.e. the sequence),
and we still win this game because we can choose devise a (probabilistic) strategy that works on all sequences. ? Denis Dec 19 '13 at 11:54
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